Math 55a: A preview of ``abstract nonsense''

Suppose we have a sequence of vector spaces and linear maps

... -->
Vn-1 -->
Vn -->
Vn+1 -->
...

The sequence is said to be ``exact at Vn''
if the kernel of the map
Vn-->Vn+1
equals the image of the map
Vn-1-->Vn.
[As a consequence, the composite map
Vn-1-->Vn+1
must be the zero map.] The sequence is said to be ``exact''
if it is exact at each vector space with an incoming and outgoing arrow.

The simplest cases:

0 --> V1 --> 0
is exact iff V1 is the zero space.

0 --> V1 --> V2 --> 0
is exact iff the map
V1-->V2
is an isomorphism.

0 --> V1 --> V2
--> V3 --> 0
is exact iff
V1-->V2
is an injection,
V2-->V3
is a surjection, and the image of the first map
equals the kernel of the second; that is, if and only if
the sequence is equivalent to

0 --> U --> V
--> V/U --> 0

for some vector space V with a subspace U
mapped into V using the restriction to U
of the identity map on V.

An exact sequence
0 --> U --> V
--> V/U --> 0
is called a ``short exact sequence''; an exact sequence
involving four or more vector spaces between the initial and final zero
is called a ``long exact sequence''. In any exact sequence
of finite-dimensional vector spaces with an initial and final zero,
the dimensions of the even- and odd-numbered vector spaces
in the sequence have the same sum; in other words,
the alternating sum of the dimensions
(a.k.a. the ``Euler characteristic'' of the sequence) vanishes.
[Check that this holds for the above cases of sequences of length
at most 3.]

(Much the same definitions are made for sequences
in some other ``categories'', such as groups,
or modules over a given ring. For instance,

0 --> Z --> Z
--> Z/NZ --> 0

is a short exact sequence of commutative groups
if we use multiplication by N as the map
Z-->Z.)

If the vector spaces Vn
in our exact sequence are finite dimensional
then the dual spaces Vn*
form an exact sequence with the arrows going in the opposite direction:

... <--
Vn-1 <--
Vn <--
Vn+1 <--
...

[A mathematician enamored with abstract nonsense
would express this fact by saying that ``duality
is an exact contravariant functor on the catogery
of finite-dimensional vector spaces and linear maps'';
the canonical identification of the second dual
V** of every finite-dimensiona
vector space V with V would likewise
give rise to an ``exact covariant functor'' on the same category.]

For instance, if we dualize

0 --> U --> V
--> V/U --> 0

we get

0 <-- U* <-- V*
<-- (V/U)* <-- 0

Which is to say that (V/U)*
is the kernel of a surjective linear map from
V* to U*
obtained from the injection of U into V.
This map is none other than the restriction map from
linear functionals on V to linear functionals on U;
the kernel of this map consists exactly of those linear functionals
that vanish on all of U. So we recover the fact that
(V/U)* is the annihilator
of U in V*,
and that the quotient of V*
by its subspace (V/U)*
is canonically identified with U*.

How much of this still works if we drop the requirement that
V be finite dimensional?